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Anettt [7]
1 year ago
6

Find the potential solution to the equation log4(2 – x) = log4(–5x – 18).

Mathematics
2 answers:
NNADVOKAT [17]1 year ago
4 0
Log₄(2 – x) = log₄(–5x – 18).

Since the logs have the same basis (4) & are equal then:

(2 – x) = (–5x – 18).  . Now solve for x & you will find x = -5
USPshnik [31]1 year ago
4 0

the answer for this equation is: -5

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restaurant etiquette dictates that you should leave a 15% tip for the server if the service is acceptable. If that is the case,
masha68 [24]
3.60
bc using a part to whole chart you cross multiply 24*15 get 360 then divide it by 100 and get 3.60





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1 year ago
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Which expression shows a way to find 20% of 950?<br> 950<br> 20<br> 20<br> What is it
Komok [63]
To find 20% of 950 you would set it up as a proportion. When doing percentages the prevent is always out of 100 so the first step would be 20/100. You are trying to find a number out of 950 so the second part would be ?/950. Now you want to cross multiply and divide. 20*950=19,000 then you divide it by 100 (your other number) 19,000/100=190. So 20% of 950 is 190.
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1 year ago
Leo is testing different types of greenhouse material to determine which type is most effective for growing strawberry bushes. H
alexandr1967 [171]

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7 0
2 years ago
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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
1 year ago
−15x+4≤109 OR −6x+70&gt;−2
Xelga [282]

-15x+4\leq109\qquad\text{subtract 4 from both sides}\\\\-15x\leq105\qquad\text{change the signs}\\\\15x\geq-105\qquad\text{divide both sides by 15}\\\\\boxed{x\geq-7}\\or\\-6x+70>-2\qquad\text{subtract 70 from both sides}\\\\-6x>-72\qquad\text{change the signs}\\\\6x

3 0
1 year ago
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